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rational

2 years ago

.

UsukiDoll

2 years ago

u substitution?

UsukiDoll

2 years ago

like let u = x^2
du = 2dx
\[\frac{du}{2} = dx\]
and then -cos x
because the derivative of cos x is -sin x and if we have that extra negative we can multiply . A negative x negative is positive
-(-sinx) = sinx

so to gain sin(u) back
I need -cos(u)
because the derivative of cos(u) is -sin(u) and with that extra negative in place... a negative x a negative is a positive so -(-sin(u)) ->sin (u)
\[\frac{-1}{2x}\cos(x^2)+C\]

UsukiDoll

2 years ago

FORGET IT! It looks innocent, yet when I try to do it, it's all screwed up.

now I change variable:
\[\large {x^2} = z\]
where z is the new integration variable

anonymous

2 years ago

There is no anti-derivative for \(\sin x^2\) in terms of elementary functions. But there is a way to solve this integral with elementary functions. Also a little bit of Laplace transforms needed here and first thing to do is changing it to a double integral.

I think on one of the circles integral has to have some value, because the answer is\[\sqrt{\frac{\pi}{8}}\]

anonymous

2 years ago

Looking at it, I guess the way that made sense to me was considering \(cos(x^{2}) + isin(x^{2}) = e^{ix^{2}}\) and then dealing with the integral of \(e^{ix^{2}}\)

anonymous

2 years ago

@Concentrationalizing that will lead us to answer also, quite right

Michele_Laino

2 years ago

the function:
\[{\frac{{\sin \left( z \right)}}{{2\sqrt z }}}\]
has not poles in z-plane, it has only a singularity at z=0, nevertheless the residual value at z=0, is 0

anonymous

2 years ago

I wish I perfectly understand this process, but I do know a little bit to type up a jumbled mess. So first note this:
\(cos(x^{2}) + isin(x^{2}) = e^{ix^{2}}\)
So instead of looking directly at the integral of \(sin(x^{2})\), we would want to try and mess with \(e^{ix^{2}}\).
Okay, so \(e^{iz^{2}}\) is entire, which means \(\int_C e^{iz^{2}} = 0\) for any closed contour C. For dealing with this integral, we consider this contour
|dw:1434535791428:dw|
The reason the angle of the chosen contour stops at \(\pi/4\) is because on the arc of the circle, \(z=re^{i \theta}\). Choosing \(\theta = \pi/4\) gives us this nice result:
\(e^{iz^{2}} = e^{i(re^{i\pi/4})^{2}} = e^{-r^{2}}\)
And this is nice because
\[\int\limits_{0}^{+\infty}e^{-x^{2}}dx = \frac{ \sqrt{\pi} }{ 2 }\]
So we can add up the integrals along the 3 curves and get a result
\[\int\limits_{0}^{+\infty}e^{iz^{2}}dz = \lim_{R \rightarrow +\infty} (\int\limits_{0}^{R}e^{ix^{2}}dx + 0 -e^{i\pi/4}\int\limits_{0}^{R}e^{-x^{2}}dx)\]
The middle part is 0 because the integral on C2 would just be 0 along that arc. As R goes to \(+\infty\), The first integral tends to the integral we're looking for while the 3rd integral is a known result.
\[\int\limits_{0}^{+\infty}e^{iz^{2}}dz = \int\limits_{0}^{+\infty}e^{ix^{2}}dx -(\frac{ 1 }{ \sqrt{2} }+i \frac{ 1 }{ \sqrt{2} })\cdot \frac{ \sqrt{\pi} }{ 2 }\]
Of course the left hand side is 0 as previously mentioned. Now we can just move the result of the 3rd integral to the other side and get a result.
\[\int\limits_{0}^{+\infty}e^{ix^{2}}dx = \frac{ \sqrt{\pi} }{ 2\sqrt{2} } + i \frac{ \sqrt{\pi} }{ 2\sqrt{2} }\]
\[\int\limits_{0}^{+\infty} \cos(x^{2})dx + i \int\limits_{0}^{+\infty}\sin(x^{2})dx = \frac{ \sqrt{\pi} }{ 2\sqrt{2} } + i \frac{ \sqrt{\pi} }{ 2\sqrt{2} }\]
Real parts equal real parts, imaginary parts equal imaginary parts, and we get those both the required integral as well as the one for \(\int_{0}^{+\infty}cos(x^{2})dx\) give us \(\frac{{\sqrt{\pi}}}{2\sqrt{2}}\)

I actually posted the same question a while back, and @Michele_Laino and I arrived at the answer (more or less) in much the same way.
The Laplace transform is a neat way of approaching this, provided we can accept the fact that \(\mathcal{L}\left\{x^{-1/2}\right\}=\sqrt{\dfrac{\pi}{s}}\).

anonymous

2 years ago

Actually, we used the same method as @Concentrationalizing though we through around the idea of using a contour.